Integrand size = 15, antiderivative size = 202 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=-\frac {a^{15}}{x}-\frac {45 a^{14} b}{2 x^{2/3}}-\frac {315 a^{13} b^2}{\sqrt [3]{x}}+4095 a^{11} b^4 \sqrt [3]{x}+\frac {9009}{2} a^{10} b^5 x^{2/3}+5005 a^9 b^6 x+\frac {19305}{4} a^8 b^7 x^{4/3}+3861 a^7 b^8 x^{5/3}+\frac {5005}{2} a^6 b^9 x^2+1287 a^5 b^{10} x^{7/3}+\frac {4095}{8} a^4 b^{11} x^{8/3}+\frac {455}{3} a^3 b^{12} x^3+\frac {63}{2} a^2 b^{13} x^{10/3}+\frac {45}{11} a b^{14} x^{11/3}+\frac {b^{15} x^4}{4}+455 a^{12} b^3 \log (x) \]
-a^15/x-45/2*a^14*b/x^(2/3)-315*a^13*b^2/x^(1/3)+4095*a^11*b^4*x^(1/3)+900 9/2*a^10*b^5*x^(2/3)+5005*a^9*b^6*x+19305/4*a^8*b^7*x^(4/3)+3861*a^7*b^8*x ^(5/3)+5005/2*a^6*b^9*x^2+1287*a^5*b^10*x^(7/3)+4095/8*a^4*b^11*x^(8/3)+45 5/3*a^3*b^12*x^3+63/2*a^2*b^13*x^(10/3)+45/11*a*b^14*x^(11/3)+1/4*b^15*x^4 +455*a^12*b^3*ln(x)
Time = 0.11 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=\frac {-264 a^{15}-5940 a^{14} b \sqrt [3]{x}-83160 a^{13} b^2 x^{2/3}+1081080 a^{11} b^4 x^{4/3}+1189188 a^{10} b^5 x^{5/3}+1321320 a^9 b^6 x^2+1274130 a^8 b^7 x^{7/3}+1019304 a^7 b^8 x^{8/3}+660660 a^6 b^9 x^3+339768 a^5 b^{10} x^{10/3}+135135 a^4 b^{11} x^{11/3}+40040 a^3 b^{12} x^4+8316 a^2 b^{13} x^{13/3}+1080 a b^{14} x^{14/3}+66 b^{15} x^5}{264 x}+1365 a^{12} b^3 \log \left (\sqrt [3]{x}\right ) \]
(-264*a^15 - 5940*a^14*b*x^(1/3) - 83160*a^13*b^2*x^(2/3) + 1081080*a^11*b ^4*x^(4/3) + 1189188*a^10*b^5*x^(5/3) + 1321320*a^9*b^6*x^2 + 1274130*a^8* b^7*x^(7/3) + 1019304*a^7*b^8*x^(8/3) + 660660*a^6*b^9*x^3 + 339768*a^5*b^ 10*x^(10/3) + 135135*a^4*b^11*x^(11/3) + 40040*a^3*b^12*x^4 + 8316*a^2*b^1 3*x^(13/3) + 1080*a*b^14*x^(14/3) + 66*b^15*x^5)/(264*x) + 1365*a^12*b^3*L og[x^(1/3)]
Time = 0.34 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {798, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle 3 \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^{4/3}}d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle 3 \int \left (\frac {a^{15}}{x^{4/3}}+\frac {15 b a^{14}}{x}+\frac {105 b^2 a^{13}}{x^{2/3}}+\frac {455 b^3 a^{12}}{\sqrt [3]{x}}+1365 b^4 a^{11}+3003 b^5 \sqrt [3]{x} a^{10}+5005 b^6 x^{2/3} a^9+6435 b^7 x a^8+6435 b^8 x^{4/3} a^7+5005 b^9 x^{5/3} a^6+3003 b^{10} x^2 a^5+1365 b^{11} x^{7/3} a^4+455 b^{12} x^{8/3} a^3+105 b^{13} x^3 a^2+15 b^{14} x^{10/3} a+b^{15} x^{11/3}\right )d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (-\frac {a^{15}}{3 x}-\frac {15 a^{14} b}{2 x^{2/3}}-\frac {105 a^{13} b^2}{\sqrt [3]{x}}+455 a^{12} b^3 \log \left (\sqrt [3]{x}\right )+1365 a^{11} b^4 \sqrt [3]{x}+\frac {3003}{2} a^{10} b^5 x^{2/3}+\frac {5005}{3} a^9 b^6 x+\frac {6435}{4} a^8 b^7 x^{4/3}+1287 a^7 b^8 x^{5/3}+\frac {5005}{6} a^6 b^9 x^2+429 a^5 b^{10} x^{7/3}+\frac {1365}{8} a^4 b^{11} x^{8/3}+\frac {455}{9} a^3 b^{12} x^3+\frac {21}{2} a^2 b^{13} x^{10/3}+\frac {15}{11} a b^{14} x^{11/3}+\frac {b^{15} x^4}{12}\right )\) |
3*(-1/3*a^15/x - (15*a^14*b)/(2*x^(2/3)) - (105*a^13*b^2)/x^(1/3) + 1365*a ^11*b^4*x^(1/3) + (3003*a^10*b^5*x^(2/3))/2 + (5005*a^9*b^6*x)/3 + (6435*a ^8*b^7*x^(4/3))/4 + 1287*a^7*b^8*x^(5/3) + (5005*a^6*b^9*x^2)/6 + 429*a^5* b^10*x^(7/3) + (1365*a^4*b^11*x^(8/3))/8 + (455*a^3*b^12*x^3)/9 + (21*a^2* b^13*x^(10/3))/2 + (15*a*b^14*x^(11/3))/11 + (b^15*x^4)/12 + 455*a^12*b^3* Log[x^(1/3)])
3.24.46.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 3.43 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(-\frac {a^{15}}{x}-\frac {45 a^{14} b}{2 x^{\frac {2}{3}}}-\frac {315 a^{13} b^{2}}{x^{\frac {1}{3}}}+4095 a^{11} b^{4} x^{\frac {1}{3}}+\frac {9009 a^{10} b^{5} x^{\frac {2}{3}}}{2}+5005 a^{9} b^{6} x +\frac {19305 a^{8} b^{7} x^{\frac {4}{3}}}{4}+3861 a^{7} b^{8} x^{\frac {5}{3}}+\frac {5005 a^{6} b^{9} x^{2}}{2}+1287 a^{5} b^{10} x^{\frac {7}{3}}+\frac {4095 a^{4} b^{11} x^{\frac {8}{3}}}{8}+\frac {455 a^{3} b^{12} x^{3}}{3}+\frac {63 a^{2} b^{13} x^{\frac {10}{3}}}{2}+\frac {45 a \,b^{14} x^{\frac {11}{3}}}{11}+\frac {b^{15} x^{4}}{4}+455 a^{12} b^{3} \ln \left (x \right )\) | \(165\) |
| default | \(-\frac {a^{15}}{x}-\frac {45 a^{14} b}{2 x^{\frac {2}{3}}}-\frac {315 a^{13} b^{2}}{x^{\frac {1}{3}}}+4095 a^{11} b^{4} x^{\frac {1}{3}}+\frac {9009 a^{10} b^{5} x^{\frac {2}{3}}}{2}+5005 a^{9} b^{6} x +\frac {19305 a^{8} b^{7} x^{\frac {4}{3}}}{4}+3861 a^{7} b^{8} x^{\frac {5}{3}}+\frac {5005 a^{6} b^{9} x^{2}}{2}+1287 a^{5} b^{10} x^{\frac {7}{3}}+\frac {4095 a^{4} b^{11} x^{\frac {8}{3}}}{8}+\frac {455 a^{3} b^{12} x^{3}}{3}+\frac {63 a^{2} b^{13} x^{\frac {10}{3}}}{2}+\frac {45 a \,b^{14} x^{\frac {11}{3}}}{11}+\frac {b^{15} x^{4}}{4}+455 a^{12} b^{3} \ln \left (x \right )\) | \(165\) |
| trager | \(\frac {\left (-1+x \right ) \left (3 b^{15} x^{4}+1820 a^{3} b^{12} x^{3}+3 b^{15} x^{3}+30030 a^{6} b^{9} x^{2}+1820 a^{3} b^{12} x^{2}+3 b^{15} x^{2}+60060 a^{9} b^{6} x +30030 a^{6} b^{9} x +1820 a^{3} b^{12} x +3 b^{15} x +12 a^{15}\right )}{12 x}-\frac {9 \left (-14 b^{12} x^{4}-572 a^{3} b^{9} x^{3}-2145 a^{6} b^{6} x^{2}-1820 a^{9} b^{3} x +10 a^{12}\right ) a^{2} b}{4 x^{\frac {2}{3}}}-\frac {9 \left (-40 b^{12} x^{4}-5005 a^{3} b^{9} x^{3}-37752 a^{6} b^{6} x^{2}-44044 a^{9} b^{3} x +3080 a^{12}\right ) a \,b^{2}}{88 x^{\frac {1}{3}}}+455 a^{12} b^{3} \ln \left (x \right )\) | \(224\) |
-a^15/x-45/2*a^14*b/x^(2/3)-315*a^13*b^2/x^(1/3)+4095*a^11*b^4*x^(1/3)+900 9/2*a^10*b^5*x^(2/3)+5005*a^9*b^6*x+19305/4*a^8*b^7*x^(4/3)+3861*a^7*b^8*x ^(5/3)+5005/2*a^6*b^9*x^2+1287*a^5*b^10*x^(7/3)+4095/8*a^4*b^11*x^(8/3)+45 5/3*a^3*b^12*x^3+63/2*a^2*b^13*x^(10/3)+45/11*a*b^14*x^(11/3)+1/4*b^15*x^4 +455*a^12*b^3*ln(x)
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=\frac {66 \, b^{15} x^{5} + 40040 \, a^{3} b^{12} x^{4} + 660660 \, a^{6} b^{9} x^{3} + 1321320 \, a^{9} b^{6} x^{2} + 360360 \, a^{12} b^{3} x \log \left (x^{\frac {1}{3}}\right ) - 264 \, a^{15} + 27 \, {\left (40 \, a b^{14} x^{4} + 5005 \, a^{4} b^{11} x^{3} + 37752 \, a^{7} b^{8} x^{2} + 44044 \, a^{10} b^{5} x - 3080 \, a^{13} b^{2}\right )} x^{\frac {2}{3}} + 594 \, {\left (14 \, a^{2} b^{13} x^{4} + 572 \, a^{5} b^{10} x^{3} + 2145 \, a^{8} b^{7} x^{2} + 1820 \, a^{11} b^{4} x - 10 \, a^{14} b\right )} x^{\frac {1}{3}}}{264 \, x} \]
1/264*(66*b^15*x^5 + 40040*a^3*b^12*x^4 + 660660*a^6*b^9*x^3 + 1321320*a^9 *b^6*x^2 + 360360*a^12*b^3*x*log(x^(1/3)) - 264*a^15 + 27*(40*a*b^14*x^4 + 5005*a^4*b^11*x^3 + 37752*a^7*b^8*x^2 + 44044*a^10*b^5*x - 3080*a^13*b^2) *x^(2/3) + 594*(14*a^2*b^13*x^4 + 572*a^5*b^10*x^3 + 2145*a^8*b^7*x^2 + 18 20*a^11*b^4*x - 10*a^14*b)*x^(1/3))/x
Time = 0.48 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=- \frac {a^{15}}{x} - \frac {45 a^{14} b}{2 x^{\frac {2}{3}}} - \frac {315 a^{13} b^{2}}{\sqrt [3]{x}} + 455 a^{12} b^{3} \log {\left (x \right )} + 4095 a^{11} b^{4} \sqrt [3]{x} + \frac {9009 a^{10} b^{5} x^{\frac {2}{3}}}{2} + 5005 a^{9} b^{6} x + \frac {19305 a^{8} b^{7} x^{\frac {4}{3}}}{4} + 3861 a^{7} b^{8} x^{\frac {5}{3}} + \frac {5005 a^{6} b^{9} x^{2}}{2} + 1287 a^{5} b^{10} x^{\frac {7}{3}} + \frac {4095 a^{4} b^{11} x^{\frac {8}{3}}}{8} + \frac {455 a^{3} b^{12} x^{3}}{3} + \frac {63 a^{2} b^{13} x^{\frac {10}{3}}}{2} + \frac {45 a b^{14} x^{\frac {11}{3}}}{11} + \frac {b^{15} x^{4}}{4} \]
-a**15/x - 45*a**14*b/(2*x**(2/3)) - 315*a**13*b**2/x**(1/3) + 455*a**12*b **3*log(x) + 4095*a**11*b**4*x**(1/3) + 9009*a**10*b**5*x**(2/3)/2 + 5005* a**9*b**6*x + 19305*a**8*b**7*x**(4/3)/4 + 3861*a**7*b**8*x**(5/3) + 5005* a**6*b**9*x**2/2 + 1287*a**5*b**10*x**(7/3) + 4095*a**4*b**11*x**(8/3)/8 + 455*a**3*b**12*x**3/3 + 63*a**2*b**13*x**(10/3)/2 + 45*a*b**14*x**(11/3)/ 11 + b**15*x**4/4
Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=\frac {1}{4} \, b^{15} x^{4} + \frac {45}{11} \, a b^{14} x^{\frac {11}{3}} + \frac {63}{2} \, a^{2} b^{13} x^{\frac {10}{3}} + \frac {455}{3} \, a^{3} b^{12} x^{3} + \frac {4095}{8} \, a^{4} b^{11} x^{\frac {8}{3}} + 1287 \, a^{5} b^{10} x^{\frac {7}{3}} + \frac {5005}{2} \, a^{6} b^{9} x^{2} + 3861 \, a^{7} b^{8} x^{\frac {5}{3}} + \frac {19305}{4} \, a^{8} b^{7} x^{\frac {4}{3}} + 5005 \, a^{9} b^{6} x + 455 \, a^{12} b^{3} \log \left (x\right ) + \frac {9009}{2} \, a^{10} b^{5} x^{\frac {2}{3}} + 4095 \, a^{11} b^{4} x^{\frac {1}{3}} - \frac {630 \, a^{13} b^{2} x^{\frac {2}{3}} + 45 \, a^{14} b x^{\frac {1}{3}} + 2 \, a^{15}}{2 \, x} \]
1/4*b^15*x^4 + 45/11*a*b^14*x^(11/3) + 63/2*a^2*b^13*x^(10/3) + 455/3*a^3* b^12*x^3 + 4095/8*a^4*b^11*x^(8/3) + 1287*a^5*b^10*x^(7/3) + 5005/2*a^6*b^ 9*x^2 + 3861*a^7*b^8*x^(5/3) + 19305/4*a^8*b^7*x^(4/3) + 5005*a^9*b^6*x + 455*a^12*b^3*log(x) + 9009/2*a^10*b^5*x^(2/3) + 4095*a^11*b^4*x^(1/3) - 1/ 2*(630*a^13*b^2*x^(2/3) + 45*a^14*b*x^(1/3) + 2*a^15)/x
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=\frac {1}{4} \, b^{15} x^{4} + \frac {45}{11} \, a b^{14} x^{\frac {11}{3}} + \frac {63}{2} \, a^{2} b^{13} x^{\frac {10}{3}} + \frac {455}{3} \, a^{3} b^{12} x^{3} + \frac {4095}{8} \, a^{4} b^{11} x^{\frac {8}{3}} + 1287 \, a^{5} b^{10} x^{\frac {7}{3}} + \frac {5005}{2} \, a^{6} b^{9} x^{2} + 3861 \, a^{7} b^{8} x^{\frac {5}{3}} + \frac {19305}{4} \, a^{8} b^{7} x^{\frac {4}{3}} + 5005 \, a^{9} b^{6} x + 455 \, a^{12} b^{3} \log \left ({\left | x \right |}\right ) + \frac {9009}{2} \, a^{10} b^{5} x^{\frac {2}{3}} + 4095 \, a^{11} b^{4} x^{\frac {1}{3}} - \frac {630 \, a^{13} b^{2} x^{\frac {2}{3}} + 45 \, a^{14} b x^{\frac {1}{3}} + 2 \, a^{15}}{2 \, x} \]
1/4*b^15*x^4 + 45/11*a*b^14*x^(11/3) + 63/2*a^2*b^13*x^(10/3) + 455/3*a^3* b^12*x^3 + 4095/8*a^4*b^11*x^(8/3) + 1287*a^5*b^10*x^(7/3) + 5005/2*a^6*b^ 9*x^2 + 3861*a^7*b^8*x^(5/3) + 19305/4*a^8*b^7*x^(4/3) + 5005*a^9*b^6*x + 455*a^12*b^3*log(abs(x)) + 9009/2*a^10*b^5*x^(2/3) + 4095*a^11*b^4*x^(1/3) - 1/2*(630*a^13*b^2*x^(2/3) + 45*a^14*b*x^(1/3) + 2*a^15)/x
Time = 5.97 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sqrt [3]{x}\right )^{15}}{x^2} \, dx=\frac {b^{15}\,x^4}{4}-\frac {a^{15}+\frac {45\,a^{14}\,b\,x^{1/3}}{2}+315\,a^{13}\,b^2\,x^{2/3}}{x}+1365\,a^{12}\,b^3\,\ln \left (x^{1/3}\right )+5005\,a^9\,b^6\,x+\frac {45\,a\,b^{14}\,x^{11/3}}{11}+\frac {5005\,a^6\,b^9\,x^2}{2}+\frac {455\,a^3\,b^{12}\,x^3}{3}+4095\,a^{11}\,b^4\,x^{1/3}+\frac {9009\,a^{10}\,b^5\,x^{2/3}}{2}+\frac {19305\,a^8\,b^7\,x^{4/3}}{4}+3861\,a^7\,b^8\,x^{5/3}+1287\,a^5\,b^{10}\,x^{7/3}+\frac {4095\,a^4\,b^{11}\,x^{8/3}}{8}+\frac {63\,a^2\,b^{13}\,x^{10/3}}{2} \]
(b^15*x^4)/4 - (a^15 + (45*a^14*b*x^(1/3))/2 + 315*a^13*b^2*x^(2/3))/x + 1 365*a^12*b^3*log(x^(1/3)) + 5005*a^9*b^6*x + (45*a*b^14*x^(11/3))/11 + (50 05*a^6*b^9*x^2)/2 + (455*a^3*b^12*x^3)/3 + 4095*a^11*b^4*x^(1/3) + (9009*a ^10*b^5*x^(2/3))/2 + (19305*a^8*b^7*x^(4/3))/4 + 3861*a^7*b^8*x^(5/3) + 12 87*a^5*b^10*x^(7/3) + (4095*a^4*b^11*x^(8/3))/8 + (63*a^2*b^13*x^(10/3))/2